Returning to original variable in trig substitution I have no idea how to do this part in a problem. For example if I have $$\int \frac { \sqrt{9-x^2}}{x^2}dx$$
like in my book, they make $x = 3\sin t$ and then $dx = 3\cos t\, dt$
Which is fine I guess, pretty abstract but I can work through it up until the end where they get $$-\cot t - t + c$$
Now they want to introduce $x$ back into the function and I have no idea what to do, there is a picture of a triangle which I do not really get and then some other stuff that doesn't make sense at all (all of it). The answer involves an $\arcsin$, what do I do?
 A: Look what you stated in your post.  We know 
$$x=3\sin t$$
so 
$$\arcsin \left(\frac{x}{3}\right)=t$$

This may make things a bit clearer (hopefully!).  Say we have
$$f(x)=\sqrt{1-x^2}$$
We introduce a variable $t$ that suffices $x=\sin t$.  In other words, you are assigning $t=\arcsin x$!
Thus 
$$f(x)=\sqrt{1-\sin^2 t}=\cos t$$
Now you have the function in terms of $x$ and $t$ - but it is simpler when in terms of $t$!
In more general terms, if we have $f(x)$ and $t=g^{-1}(x) \implies x = g(t)$ then $f(g(t))=f(x)$

You will note that if we plug $t=\arcsin x$ back in:
$$\cos t=\cos (\arcsin x)=\sqrt{1-x^2}$$
An easy way to see why this is is to have a triangle with angle $A=\arcsin x \implies \sin A = x$.
We know 
$$\sin A = x = x/1 = \text{opp/hyp}$$
so label the opposite side $x$ and the hypotenuse $1$.  Find the third side with Pythagorean theorem, and it is $\sqrt{1^2+x^2}=\sqrt{1+x^2}$.  From elementary trig, we have 
$$\cos A = \text{adj/hyp}=\frac{\sqrt{1+x^2}}{1}=\cos (\arcsin x) $$
